The given function is:

$F(x) = 1 - 2x^2$

This is a quadratic function where:

• The constant term is $1$.
• The coefficient of $x^2$ is $-2$, which means the parabola opens downward (since the coefficient of $x^2$ is negative).

Would you like to analyze this function further, like finding its vertex, graphing it, or solving for specific values of $F(x)$?

Let me know if you have any questions, and here are some related questions for you:

1. What is the vertex of the quadratic function $F(x) = 1 - 2x^2$?
2. How would you graph $F(x) = 1 - 2x^2$?
3. What is the value of $F(x)$ when $x = 0$?
4. How does the coefficient of $x^2$ affect the graph of a quadratic function?
5. How would you find the x-intercepts (if any) of the function?

Tip: The vertex of a quadratic function $ax^2 + bx + c$ can be found using the formula $x = \frac{-b}{2a}$.